Mixing
Velocity of a particle on a Parabola
Clash Royale CLAN TAG#URR8PPP
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2
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A particle moves along the top of the
parabola $y^2 = 2x$ from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point $(2, 2)$.
So I isolate $y$, giving me $y=sqrt2x$. I then find the derivative of $y$, which is $1/sqrt2x$. And $sqrt2x=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.
calculus derivatives
edited 3 hours ago
Zamarion
925
asked 4 hours ago
drecinos0530
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up vote
2
down vote
favorite
A particle moves along the top of the
parabola $y^2 = 2x$ from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point $(2, 2)$.
So I isolate $y$, giving me $y=sqrt2x$. I then find the derivative of $y$, which is $1/sqrt2x$. And $sqrt2x=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.
calculus derivatives
edited 3 hours ago
Zamarion
925
asked 4 hours ago
drecinos0530
303
New contributor
drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think you need to use that speed is 5, and also it moves along the parabola which has varying nearby distances at different points.
– coffeemath
4 hours ago
Find the direction of the tangent at $(2,2)$ and create a vector with that direction and length equal to $5$.
– amd
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A particle moves along the top of the
parabola $y^2 = 2x$ from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point $(2, 2)$.
So I isolate $y$, giving me $y=sqrt2x$. I then find the derivative of $y$, which is $1/sqrt2x$. And $sqrt2x=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.
calculus derivatives
edited 3 hours ago
Zamarion
925
asked 4 hours ago
drecinos0530
303
New contributor
drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A particle moves along the top of the
parabola $y^2 = 2x$ from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point $(2, 2)$.
So I isolate $y$, giving me $y=sqrt2x$. I then find the derivative of $y$, which is $1/sqrt2x$. And $sqrt2x=y$ so the derivative of also equal to $1/y$. At $(2,2)$ the derivative is 0.5. Not sure where to go from there though. Any help is appreciated.
calculus derivatives
calculus derivatives
edited 3 hours ago
Zamarion
925
asked 4 hours ago
drecinos0530
303
New contributor
drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago
Zamarion
925
asked 4 hours ago
drecinos0530
303
New contributor
drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 3 hours ago
Zamarion
925
edited 3 hours ago
Zamarion
925
edited 3 hours ago
Zamarion
925
925
asked 4 hours ago
drecinos0530
303
New contributor
drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
drecinos0530
303
asked 4 hours ago
drecinos0530
303
303
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drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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drecinos0530 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think you need to use that speed is 5, and also it moves along the parabola which has varying nearby distances at different points.
– coffeemath
4 hours ago
Find the direction of the tangent at $(2,2)$ and create a vector with that direction and length equal to $5$.
– amd
3 hours ago
add a comment |Â
I think you need to use that speed is 5, and also it moves along the parabola which has varying nearby distances at different points.
– coffeemath
4 hours ago
Find the direction of the tangent at $(2,2)$ and create a vector with that direction and length equal to $5$.
– amd
3 hours ago
I think you need to use that speed is 5, and also it moves along the parabola which has varying nearby distances at different points.
– coffeemath
4 hours ago
I think you need to use that speed is 5, and also it moves along the parabola which has varying nearby distances at different points.
– coffeemath
4 hours ago
Find the direction of the tangent at $(2,2)$ and create a vector with that direction and length equal to $5$.
– amd
3 hours ago
Find the direction of the tangent at $(2,2)$ and create a vector with that direction and length equal to $5$.
– amd
3 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
$(2y) dy/dx=2$; $dy/dx =1/y$.
Slope at $(2,2)$: $dy/dx =1/2= tan alpha$.
$cos alpha =2/√5$, $sin alpha =1/√5.$
(Pythagoras).
$v_x= v cos alpha$; $v_y = v sin alpha$, where $v = |vec v|$ .
$vec v = (v_x,v_y) $.
answered 3 hours ago
Peter Szilas
8,7002619
add a comment |Â
up vote
2
down vote
Consider the curve in a parametric form $vecz=(x(t),y(t))$, in that case we know that
$$
vecv=fracmathrmdveczmathrmdt=(x'(t),y'(t))
$$
Additionally, we know that the speed is $|vecv|=5$. Therefore, we need to find the direction of $vecv$. Given the equation of the curve
$$
y^2=2x
$$
we can write the parametric from in terms of $y$ alone:
$$
vecz=left(fracy(t)^22,y(t) right)
$$
Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then:
$$
vecz=left(fracc^2t^22,ct right)
$$
Then the velocity is
$$
vecv=(c^2t,c)
$$
At the point $vecz=(2,2)$, $t=2/c$. Therefore,
$$
vecv=(2c,c) Rightarrow |vecv|=sqrt4c^2+c^2=5
$$
Then, $c=sqrt5$. Lets check our solution,
$$
beginalign
vecz(t)=&left(frac5t^22,sqrt5tright)\
vecv(t)=&left(5t,sqrt5right)
endalign
$$
at $t=frac2sqrt5$:
$$
beginalign
veczleft(frac2sqrt5right)=&left(frac52cdotfrac45,sqrt5frac2sqrt5right)=(2,2)\
vecvleft(frac2sqrt5right)=&left(5frac2sqrt5,sqrt5right)=(2sqrt5,sqrt5)
endalign
$$
answered 3 hours ago
PabloG.
762
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up vote
2
down vote
In that case is useful to use parametric equation that is
$$y^2=2x implies vec s(t)=(2c^2t^2,2ct) implies vec v(t)=s'(t)=(4c^2t,2c)$$
and since the point $(2,2)$ is reached at $t=1/c$, that is $vec s(1/c)=(2,2)$, the velocity at that point is $vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have
$$|v(1/c)|=sqrt16c^2+4c^2=5 implies c=fracsqrt 52$$
therefore the velocity at $(2,2)$ is
$$v(t)=left(2sqrt 5,sqrt 5right)$$
As an alternative
$$y^2=2x implies 2ydy=2dx implies fracdydx=frac1y$$
therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find
$$5=ksqrt16+4implies k=fracsqrt 52$$
answered 3 hours ago
gimusi
75.8k73889
add a comment |Â
up vote
0
down vote
in cartesian co-ordinate sysytem :
$(speed)^2=left(dfracdxdtright)^2+left(dfracdydtright)^2=25$ also,
$left(dfracdydtright)=dfrac1yleft(dfracdxdtright) $ put it in above equation
to get ,
$left(dfracdxdtright)=dfrac5ysqrt1+y^2$ and $left(dfracdydtright)=dfrac5sqrt1+y^2$
thus, at point (2,2) speed will be
$sqrt (2sqrt5)^2+(sqrt 5)^2=5 $ unit per second
in above calculations gravity is not taken into account
answered 3 hours ago
veeresh pandey
758314
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$(2y) dy/dx=2$; $dy/dx =1/y$.
Slope at $(2,2)$: $dy/dx =1/2= tan alpha$.
$cos alpha =2/√5$, $sin alpha =1/√5.$
(Pythagoras).
$v_x= v cos alpha$; $v_y = v sin alpha$, where $v = |vec v|$ .
$vec v = (v_x,v_y) $.
answered 3 hours ago
Peter Szilas
8,7002619
add a comment |Â
up vote
3
down vote
$(2y) dy/dx=2$; $dy/dx =1/y$.
Slope at $(2,2)$: $dy/dx =1/2= tan alpha$.
$cos alpha =2/√5$, $sin alpha =1/√5.$
(Pythagoras).
$v_x= v cos alpha$; $v_y = v sin alpha$, where $v = |vec v|$ .
$vec v = (v_x,v_y) $.
answered 3 hours ago
Peter Szilas
8,7002619
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$(2y) dy/dx=2$; $dy/dx =1/y$.
Slope at $(2,2)$: $dy/dx =1/2= tan alpha$.
$cos alpha =2/√5$, $sin alpha =1/√5.$
(Pythagoras).
$v_x= v cos alpha$; $v_y = v sin alpha$, where $v = |vec v|$ .
$vec v = (v_x,v_y) $.
answered 3 hours ago
Peter Szilas
8,7002619
$(2y) dy/dx=2$; $dy/dx =1/y$.
Slope at $(2,2)$: $dy/dx =1/2= tan alpha$.
$cos alpha =2/√5$, $sin alpha =1/√5.$
(Pythagoras).
$v_x= v cos alpha$; $v_y = v sin alpha$, where $v = |vec v|$ .
$vec v = (v_x,v_y) $.
answered 3 hours ago
Peter Szilas
8,7002619
answered 3 hours ago
Peter Szilas
8,7002619
answered 3 hours ago
Peter Szilas
8,7002619
answered 3 hours ago
Peter Szilas
8,7002619
8,7002619
add a comment |Â
add a comment |Â
up vote
2
down vote
Consider the curve in a parametric form $vecz=(x(t),y(t))$, in that case we know that
$$
vecv=fracmathrmdveczmathrmdt=(x'(t),y'(t))
$$
Additionally, we know that the speed is $|vecv|=5$. Therefore, we need to find the direction of $vecv$. Given the equation of the curve
$$
y^2=2x
$$
we can write the parametric from in terms of $y$ alone:
$$
vecz=left(fracy(t)^22,y(t) right)
$$
Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then:
$$
vecz=left(fracc^2t^22,ct right)
$$
Then the velocity is
$$
vecv=(c^2t,c)
$$
At the point $vecz=(2,2)$, $t=2/c$. Therefore,
$$
vecv=(2c,c) Rightarrow |vecv|=sqrt4c^2+c^2=5
$$
Then, $c=sqrt5$. Lets check our solution,
$$
beginalign
vecz(t)=&left(frac5t^22,sqrt5tright)\
vecv(t)=&left(5t,sqrt5right)
endalign
$$
at $t=frac2sqrt5$:
$$
beginalign
veczleft(frac2sqrt5right)=&left(frac52cdotfrac45,sqrt5frac2sqrt5right)=(2,2)\
vecvleft(frac2sqrt5right)=&left(5frac2sqrt5,sqrt5right)=(2sqrt5,sqrt5)
endalign
$$
answered 3 hours ago
PabloG.
762
New contributor
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
Consider the curve in a parametric form $vecz=(x(t),y(t))$, in that case we know that
$$
vecv=fracmathrmdveczmathrmdt=(x'(t),y'(t))
$$
Additionally, we know that the speed is $|vecv|=5$. Therefore, we need to find the direction of $vecv$. Given the equation of the curve
$$
y^2=2x
$$
we can write the parametric from in terms of $y$ alone:
$$
vecz=left(fracy(t)^22,y(t) right)
$$
Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then:
$$
vecz=left(fracc^2t^22,ct right)
$$
Then the velocity is
$$
vecv=(c^2t,c)
$$
At the point $vecz=(2,2)$, $t=2/c$. Therefore,
$$
vecv=(2c,c) Rightarrow |vecv|=sqrt4c^2+c^2=5
$$
Then, $c=sqrt5$. Lets check our solution,
$$
beginalign
vecz(t)=&left(frac5t^22,sqrt5tright)\
vecv(t)=&left(5t,sqrt5right)
endalign
$$
at $t=frac2sqrt5$:
$$
beginalign
veczleft(frac2sqrt5right)=&left(frac52cdotfrac45,sqrt5frac2sqrt5right)=(2,2)\
vecvleft(frac2sqrt5right)=&left(5frac2sqrt5,sqrt5right)=(2sqrt5,sqrt5)
endalign
$$
answered 3 hours ago
PabloG.
762
New contributor
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Consider the curve in a parametric form $vecz=(x(t),y(t))$, in that case we know that
$$
vecv=fracmathrmdveczmathrmdt=(x'(t),y'(t))
$$
Additionally, we know that the speed is $|vecv|=5$. Therefore, we need to find the direction of $vecv$. Given the equation of the curve
$$
y^2=2x
$$
we can write the parametric from in terms of $y$ alone:
$$
vecz=left(fracy(t)^22,y(t) right)
$$
Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then:
$$
vecz=left(fracc^2t^22,ct right)
$$
Then the velocity is
$$
vecv=(c^2t,c)
$$
At the point $vecz=(2,2)$, $t=2/c$. Therefore,
$$
vecv=(2c,c) Rightarrow |vecv|=sqrt4c^2+c^2=5
$$
Then, $c=sqrt5$. Lets check our solution,
$$
beginalign
vecz(t)=&left(frac5t^22,sqrt5tright)\
vecv(t)=&left(5t,sqrt5right)
endalign
$$
at $t=frac2sqrt5$:
$$
beginalign
veczleft(frac2sqrt5right)=&left(frac52cdotfrac45,sqrt5frac2sqrt5right)=(2,2)\
vecvleft(frac2sqrt5right)=&left(5frac2sqrt5,sqrt5right)=(2sqrt5,sqrt5)
endalign
$$
answered 3 hours ago
PabloG.
762
New contributor
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the curve in a parametric form $vecz=(x(t),y(t))$, in that case we know that
$$
vecv=fracmathrmdveczmathrmdt=(x'(t),y'(t))
$$
Additionally, we know that the speed is $|vecv|=5$. Therefore, we need to find the direction of $vecv$. Given the equation of the curve
$$
y^2=2x
$$
we can write the parametric from in terms of $y$ alone:
$$
vecz=left(fracy(t)^22,y(t) right)
$$
Moreover, we can pick the parameter $t$ such that $y(t)=ct$, where $c$ is a constant. Then:
$$
vecz=left(fracc^2t^22,ct right)
$$
Then the velocity is
$$
vecv=(c^2t,c)
$$
At the point $vecz=(2,2)$, $t=2/c$. Therefore,
$$
vecv=(2c,c) Rightarrow |vecv|=sqrt4c^2+c^2=5
$$
Then, $c=sqrt5$. Lets check our solution,
$$
beginalign
vecz(t)=&left(frac5t^22,sqrt5tright)\
vecv(t)=&left(5t,sqrt5right)
endalign
$$
at $t=frac2sqrt5$:
$$
beginalign
veczleft(frac2sqrt5right)=&left(frac52cdotfrac45,sqrt5frac2sqrt5right)=(2,2)\
vecvleft(frac2sqrt5right)=&left(5frac2sqrt5,sqrt5right)=(2sqrt5,sqrt5)
endalign
$$
answered 3 hours ago
PabloG.
762
New contributor
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 3 hours ago
PabloG.
762
New contributor
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
PabloG.
762
answered 3 hours ago
PabloG.
762
762
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PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
PabloG. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
2
down vote
In that case is useful to use parametric equation that is
$$y^2=2x implies vec s(t)=(2c^2t^2,2ct) implies vec v(t)=s'(t)=(4c^2t,2c)$$
and since the point $(2,2)$ is reached at $t=1/c$, that is $vec s(1/c)=(2,2)$, the velocity at that point is $vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have
$$|v(1/c)|=sqrt16c^2+4c^2=5 implies c=fracsqrt 52$$
therefore the velocity at $(2,2)$ is
$$v(t)=left(2sqrt 5,sqrt 5right)$$
As an alternative
$$y^2=2x implies 2ydy=2dx implies fracdydx=frac1y$$
therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find
$$5=ksqrt16+4implies k=fracsqrt 52$$
answered 3 hours ago
gimusi
75.8k73889
add a comment |Â
up vote
2
down vote
In that case is useful to use parametric equation that is
$$y^2=2x implies vec s(t)=(2c^2t^2,2ct) implies vec v(t)=s'(t)=(4c^2t,2c)$$
and since the point $(2,2)$ is reached at $t=1/c$, that is $vec s(1/c)=(2,2)$, the velocity at that point is $vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have
$$|v(1/c)|=sqrt16c^2+4c^2=5 implies c=fracsqrt 52$$
therefore the velocity at $(2,2)$ is
$$v(t)=left(2sqrt 5,sqrt 5right)$$
As an alternative
$$y^2=2x implies 2ydy=2dx implies fracdydx=frac1y$$
therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find
$$5=ksqrt16+4implies k=fracsqrt 52$$
answered 3 hours ago
gimusi
75.8k73889
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In that case is useful to use parametric equation that is
$$y^2=2x implies vec s(t)=(2c^2t^2,2ct) implies vec v(t)=s'(t)=(4c^2t,2c)$$
and since the point $(2,2)$ is reached at $t=1/c$, that is $vec s(1/c)=(2,2)$, the velocity at that point is $vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have
$$|v(1/c)|=sqrt16c^2+4c^2=5 implies c=fracsqrt 52$$
therefore the velocity at $(2,2)$ is
$$v(t)=left(2sqrt 5,sqrt 5right)$$
As an alternative
$$y^2=2x implies 2ydy=2dx implies fracdydx=frac1y$$
therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find
$$5=ksqrt16+4implies k=fracsqrt 52$$
answered 3 hours ago
gimusi
75.8k73889
In that case is useful to use parametric equation that is
$$y^2=2x implies vec s(t)=(2c^2t^2,2ct) implies vec v(t)=s'(t)=(4c^2t,2c)$$
and since the point $(2,2)$ is reached at $t=1/c$, that is $vec s(1/c)=(2,2)$, the velocity at that point is $vec v(1/c)=(4c,2c)$ and since the speed is $5$ we have
$$|v(1/c)|=sqrt16c^2+4c^2=5 implies c=fracsqrt 52$$
therefore the velocity at $(2,2)$ is
$$v(t)=left(2sqrt 5,sqrt 5right)$$
As an alternative
$$y^2=2x implies 2ydy=2dx implies fracdydx=frac1y$$
therefore at the point $(2,2)$ the tangent vector is $(4,2)$ and then the velocity vector is $k(4,2)$ and form the condition on the speed we find
$$5=ksqrt16+4implies k=fracsqrt 52$$
answered 3 hours ago
gimusi
75.8k73889
edited 2 hours ago
answered 3 hours ago
gimusi
75.8k73889
answered 3 hours ago
gimusi
75.8k73889
answered 3 hours ago
gimusi
75.8k73889
75.8k73889
add a comment |Â
add a comment |Â
up vote
0
down vote
in cartesian co-ordinate sysytem :
$(speed)^2=left(dfracdxdtright)^2+left(dfracdydtright)^2=25$ also,
$left(dfracdydtright)=dfrac1yleft(dfracdxdtright) $ put it in above equation
to get ,
$left(dfracdxdtright)=dfrac5ysqrt1+y^2$ and $left(dfracdydtright)=dfrac5sqrt1+y^2$
thus, at point (2,2) speed will be
$sqrt (2sqrt5)^2+(sqrt 5)^2=5 $ unit per second
in above calculations gravity is not taken into account
answered 3 hours ago
veeresh pandey
758314
add a comment |Â
up vote
0
down vote
in cartesian co-ordinate sysytem :
$(speed)^2=left(dfracdxdtright)^2+left(dfracdydtright)^2=25$ also,
$left(dfracdydtright)=dfrac1yleft(dfracdxdtright) $ put it in above equation
to get ,
$left(dfracdxdtright)=dfrac5ysqrt1+y^2$ and $left(dfracdydtright)=dfrac5sqrt1+y^2$
thus, at point (2,2) speed will be
$sqrt (2sqrt5)^2+(sqrt 5)^2=5 $ unit per second
in above calculations gravity is not taken into account
answered 3 hours ago
veeresh pandey
758314
add a comment |Â
up vote
0
down vote
up vote
0
down vote
in cartesian co-ordinate sysytem :
$(speed)^2=left(dfracdxdtright)^2+left(dfracdydtright)^2=25$ also,
$left(dfracdydtright)=dfrac1yleft(dfracdxdtright) $ put it in above equation
to get ,
$left(dfracdxdtright)=dfrac5ysqrt1+y^2$ and $left(dfracdydtright)=dfrac5sqrt1+y^2$
thus, at point (2,2) speed will be
$sqrt (2sqrt5)^2+(sqrt 5)^2=5 $ unit per second
in above calculations gravity is not taken into account
answered 3 hours ago
veeresh pandey
758314
in cartesian co-ordinate sysytem :
$(speed)^2=left(dfracdxdtright)^2+left(dfracdydtright)^2=25$ also,
$left(dfracdydtright)=dfrac1yleft(dfracdxdtright) $ put it in above equation
to get ,
$left(dfracdxdtright)=dfrac5ysqrt1+y^2$ and $left(dfracdydtright)=dfrac5sqrt1+y^2$
thus, at point (2,2) speed will be
$sqrt (2sqrt5)^2+(sqrt 5)^2=5 $ unit per second
in above calculations gravity is not taken into account
answered 3 hours ago
veeresh pandey
758314
answered 3 hours ago
veeresh pandey
758314
answered 3 hours ago
veeresh pandey
758314
answered 3 hours ago
veeresh pandey
758314
758314
add a comment |Â
add a comment |Â
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I think you need to use that speed is 5, and also it moves along the parabola which has varying nearby distances at different points.
– coffeemath
4 hours ago
Find the direction of the tangent at $(2,2)$ and create a vector with that direction and length equal to $5$.
– amd
3 hours ago